The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Answer:
D. 0.75 grams
Explanation:
The data given on the iridium 182 are;
The half life of the iridium 182, 
= 15 years
The mass of the sample of iridium, N₀ = 3 grams
The amount left, N(t) after two half lives is given as follows;
For two half lives, t = 2 ×
∴ t = 2 × 15 = 30
∴ The amount left, N(t) = 0.75 grams
Answer:
Option B. At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.
C. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.
Explanation:
Answer:
Freezing point is -2.81°C
Explanation:
34g/342gmol^-1 = 0.0994mol
n = m/mr
Molarity= 0.994/ 0.66 = 1.51M
◇T = -i × m ×Kf
Where ◇T is freezing depression
i= Vant Hoff factor
m = molarity
Kf = freezing content = 1.
860kgmol^-1
◇T =-1 × 1.51 × 1.860 = - 2.81°C
